Question

Intensity level 200 cm from a source of sound is 80 dB. If there is no loss of acoustic power in air and intensity of threshold hearing is $10^{- 12}Wm^{- 2}$ then, what is the intensity level at a distance of 400 cm from source

• A. Zero
• B. 54 dB
• C. 64 dB
• D. 44 dB

Answer 1

Answer: (B)

54 dB

Answer 2

$I \propto \frac{1}{r^{2}} \Rightarrow \frac{I_{2}}{I_{1}} = \frac{r_{1}^{2}}{r_{2}^{2}} = \frac{2^{2}}{(40)^{2}} = \frac{1}{400}$⇒ $I_{1} = 400I_{2}$

Intensity level at point 1,

$L_{1} = 10\log_{10}\left( \frac{I_{1}}{I_{0}} \right)$

and intensity at point 2,

$L_{2} = 10\log_{10}\left( \frac{I_{2}}{I_{0}} \right)$

∴ $L_{1} - L_{2} = 10\log\frac{I_{1}}{I_{2}} = 10\log_{10}(400)$

⇒ $L_{1} - L_{2} = 10 \times 2.602 = 26$

$L_{2} = L_{1} - 26 = 80 - 26 = 54\mspace{6mu} dB$